Solutions ch07.doc

8/17/2019 Solutions ch07.doc

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Solutions to Chapter 7 Problems

Section 7.1. Energy and Change

7.1. A slice of cheese pizza typically contains 180 Calories of food energy. The human

heart requires about 1 J of energy for each beat.

(a) Assuming that when you metabolize the pizza, all of the food energy will be

used to keep your heart beating, how many heart beats can this slice of pizza

sustain? About how many minutes would this keep your heart going? Hint : 1

Calorie = 1 kcal.

(b) Assume the efficiency of use of this food energy is about the same as that

calculated for capturing the energy from glucose oxidation, Section 7.9. What are

your answers for part (a) in this case? Explain your reasoning.

Answer to 7.1:

180 Cal×1 κχαλ

1 Χαλ×

4184 κϑ

1 κχαλ×

1000 ϑ

1 κϑ×

1 ηεαρτβεατ

1 ϑ=8×10

8 ηεαρτβεα

7.2. Pasta is largely starch and a deep-fried potato chip is mostly starch and fat.

(a) What do your observations from Investigate This 7.1 tell you about the

energy value of these foods? Do they combine exothermically with oxygen? What

is the evidence and reasoning for your response?

(b) Starch is a polymer of glucose. Is this composition consistent with your

answers in part (a) and other data from Investigate This 7.1? Explain your

reasoning.

Answer to 7.2:

(a) Once they have been ignited both the pasta and potato chip continue to

burn on their own. They release energy in the form of light and enough

thermal energy to keep their temperatures high enough to continue

burning. You might find that a potato chip burns somewhat better than

pasta, because the fat is a better fuel than starch. This is reflected in their

nutritional Calorie values as well. A gram of starch provides about 4

Calories (17 kJ) and a gram of fat about 9 Calories (38 kJ).

(b) The oxidation of glucose has been used as an example throughout the chapter.

You know that the oxidation is quite exothermic, which is shown by the burning

pasta, potato chip, and marshmallow (mainly sugars including glucose). It doesn’t

matter that the glucose units are strung together as a polymer in starch or

Chapter 7 ACS Chemistry Beta Draft

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separated in a marshmallow -- each still oxidizes to produce a large output of

enthalpy.

7.3. Predict how the results of Investigate This 7.1 would differ if a baked, rather than

a fried, potato chip were used. What will happen if a raw potato slice is used?What do you predict will be observed if a chip using Olestra® is used? In each

case, explain the reasons for your prediction.

Answer to 7.3:

The baked chip will still burn, but it is predicted not to burn as vigorously as the

fried chip because there is less residual fat available as fuel on the baked chip.

The raw potato slice may not burn at all because it has a high water content. The

Olestra® chip will also burn quite well, for the Olestra is a long-chain polymeric

fatty acid. It does not produce heat value in the body because it cannot be

metabolized.

Section 7.2. Thermal Energy (Heat) and Mechanical Energy (Work)

7.4. Several different types of energy have been discussed in this chapter. These

include kinetic energy, mechanical energy (work), potential energy, and thermal

energy (heat). How are these terms related?

Answer to 7.4:

Potential energy and kinetic energy are the two types of energy.

Mechanical energy and thermal energy are both examples of kineticenergy. This is a visual representation of the relationship.

mecancaener ter7.5. Figure 7.2 illustrates a model for the movement of combustion gases.

(a) Draw a similar diagram for a tethered helium-filled balloon. Hint: Unlike

Figure 7.2, there will only be one type of particle, representing helium atoms.

(b) Will the helium atoms in the tethered balloon exhibit any directed motion?

Why or why not? What will happen to the motion of the helium atoms in the

balloon when the tether is released? Explain briefly.


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Answer to 7.5:

(a) Tethered Balloon

(b) The helium atoms in the balloon are in constant, random motion. They

do not exhibit any direction motion. The helium atoms in the balloon are

still in constant, random motion, keeping the balloon fully inflated. The

balloon itself will drift upward, buoyed by the air molecules around it.

7.6. Consider a drop of water in a waterfall. At the very top of the fall, the molecules

in the drop are moving randomly. While it is falling (pulled down by gravity), its

molecules have an overall downward motion. When it hits the pool at the bottom,

the molecular motion again becomes random.

(a) The water at the bottom of the waterfall is a little warmer than the water atthe top. What can you say about the change in internal energy of your drop of

water as it goes from the top of the waterfall to the bottom?

(b) Has work been done or has thermal energy been transferred to cause the

change in the internal energy of the water drop? Explain where the work and/or

thermal energy come from.

Answer to 7.6:

(a) Since the temperature of the water increases when it falls and hits the

pool, its internal energy must have increased.(b) Work has been done on the water to increase its internal energy (and, hence,

its temperature). The work is done on the water when its potential energy at the

top of the fall is converted to kinetic (directed) motion as it is falling and then this

directed motion is changed to random motion when it hits the pool at the bottom

and rapidly decelerates due to the force of the collision.


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7.7. When heat is added to H2O to do work, why does the vapor form of water

generate more work than either ice or liquid water?

Answer to 7.7:

Water molecules in ice are tightly ordered in the crystal lattice. These

molecules are held in place by hydrogen bonds that move only slightly

within this lattice. Liquid water molecules also hydrogen bond, but are

capable of breaking and remaking these relatively weak intermolecular

attractions as these water molecules move relative to one another. Keep in

mind that the molecules of liquid water move slowly compared to water

molecules in the vapor phase. There is more potential energy than kinetic

energy associated with these two phases of water. In the vapor phase,

where water molecules are on average, quite distant from their neighbors,

hydrogen bonding is negligible. All the heat energy added to the vapor

phase water is transformed into kinetic energy and from there into useful

mechanical work. On the other hand, heat energy added to liquid water is

used primarily to break hydrogen bonds and only secondarily to increase

the average kinetic energy of the molecules.

7.8. Why is it important to you, as a consumer, for automotive engineers to design

engines that convert the highest possible percentage of energy to work?

Answer to 7.8:

Economic and environmental factors are important to every consumer. Whatever

the price of gasoline, consumers want to obtain the maximum number of miles

possible per gallon of gasoline. This can only be done if the chemical energy from

the fuel is converted efficiently as possible to work, and not wasted as heat.

Efficient conversion of fuel energy to work also means release of lower

concentrations of carbon monoxide and hydrocarbons into the atmosphere.

Section 7.3. Thermal Energy (Heat) Transfer

7.9. For each case below, is thermal energy being transferred by radiation or by a

contact process of conduction and/or convection? Briefly justify your choice.

(a) A silver spoon at room temperature, when placed in a cup of hot water,

becomes too warm to touch.

(b) An apple pie is baked in an electric oven.


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(c) The water in an outdoor swimming pool cools from 25 °C to 20 °C as the

summer season changes into autumn.

(d) You sunbathe with your back exposed to the sun, until your back feels very

warm.

Answer to 7.9:

(a) The process of thermal energy transfer is by contact , using conduction.

The water molecules are moving more rapidly, on the average, than the

silver atoms in the spoon. Thermal energy is transferred from the warmer,

more energetic water molecules to the cooler, less energetic silver atoms

in the spoon, causing it to become uncomfortably warm.

(b) Thermal energy transfer is likely taking place through contact . Both

conduction and convection play a role in baking this apple pie. The

electric heating unit will transfer thermal energy to molecules of gases in

the air directly in contact with the unit; this is conduction. The warmer air

rises, allowing the denser, cooler air to circulate near the heating coil; this

sets up a convection current. (In ovens advertised as “convection ovens”,

there is at least one fan to help distribute the air evenly. ) The molecules of

hot gas that are in contact with the pie will conduct thermal energy into the

pie, eventually resulting in baking the pie.

(c) The water temperature is going down because thermal energy is being

lost by contact between warmer water molecules and the cooler air.

Conduction is involved directly at the water/air interface, but convection

currents in both water and air are likely the major factors in this heat

transfer process.

(d) Thermal energy is being transferred from the sun to your back by means of the

long wavelength infrared electromagnetic radiation. As the radiation is absorbed

by the skin on your back, the molecules in your skin are moving faster which

increases their temperature. Infrared radiation is accompanied by the more

energetic ultraviolet rays, so remember to use sunscreen to avoid skin damage.

7.10. At the end of a skating season, an indoor ice rink was closed. Without any outside

cooling, how will the roof and the ice rink floor reach a common temperature?

Will it be a slow or fast process? Explain your answer.

Answer to 7.10:


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Since the cooler air near the ice rink floor has a higher density, it will stay

stagnant for a relatively long time. The ice rink floor will remain cool until it is

warmed up by heat conduction from the roof or the outside through the walls.

Heat will be conducted by the interactions of warm and cool air molecules. This

process is not going to be fast. Convection is very inefficient in this situation.

7.11. After being immersed in cold water, the core body temperature of an individual

may be abnormally low. Design three methods, one using convection, the second

using conduction, and the third using radiation to transfer heat to warm up an

individual as quickly and safely as possible. Which method will be the most

efficient in this case?

Answer to 7.11:

The convection method might consist of a small chamber with a heated floor.

The problem with this heating method is that a border of stagnant, cold air may

surround the individual. Since the cooler air near the individual has a higher

density, it will stay stagnant for a relatively long time and prevent efficient

warming.. The radiation method may consist of a large chamber with secured,

open flame inside. In this method, thermal energy will be transferred to the victim

from the flame by infrared electromagnetic radiation,. It is not going to be safe

and efficient method because of the risks of burning the individual. Also, since

radiation travels in straight lines, not all areas of the individual will be warmed

equally. Wrapping the victim in an electric blanket will be the most efficient

process of heating. In this method, heat is transferred by conduction. The

individual will be heated quickly and uniformly.

Section 7.4. State Functions and Path Functions

7.12. Your monthly bank statement gives the opening balance for your account, details

the transactions that have occurred, and reports the ending balance. Which aspects

of this statement correspond to state functions and which to path-dependent

functions?

Answer to 7.12:

The difference between the ending and initial balance corresponds to a

state function.

∆ Balance = Balanceend − Balancestart


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All of the transactions correspond to path-dependent functions. They describe all

the different expenditures and deposits that are responsible for the overall change

in balance.

7.13.The quarterback starts a play on his own 35 yard line, but is pushed back by thedefense to his own 25 yard line. He then successfully completes a pass to the

opponent’s 40 yard line. What is the number of yards gained in the play? Is the

number of yards gained a state function or a path-dependent function?

Answer to 7.13:

The play has moved the team from their own 35 yard line, past midfield, and to

the opponent’s 40 yard line. The team has moved the ball 25 yards closer to the

end zone, another example of a state function. The path taken does not matter,

only the fact that this is a first-down play successfully completed.

7.14. Today’s most advanced fossil-fuel burning power plants producing electricity

operate at an efficiency of about 42%. Suggest some reasons why a higher

percentage of the energy from burning the fossil fuels is not converted to work.

Do you think it is theoretically possible to convert 100% of the energy into work?

Explain your reasoning.

Answer to 7.14:

A significant fraction of the heat energy is transferred to the water used to cool

the turbines, and there are many sources of friction in the turbines and generator.Even when the electricity has been generated, there will be losses over long-

distance power lines usually required to reach the power grid. Most students will

intuitively recognize that realize this seems unlikely and the introduction to this

chapter tells them it is not possible. Heat losses must present serious limitations,

given that the best engineers have only found a 42% efficiency rate possible. A

more complete answer to this question will be possible after studying Chapter 8,

Entropy, where the Second Law of Thermodynamics is studied in more detail.

7.15. Consider two different cases in which a fully charged battery becomes totally

discharged.

Case 1: The battery is used to provide power to a flashlight.

Case 2: The battery is used to provide power for a child’s toy car.


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(a) Will the change from the battery being fully charged to being totally

discharged be a state function or a path-dependent function in each case? Explain

your reasoning.

(b) Will there be any thermal energy transfer by radiation or by contact in either

use of the battery? Explain your reasoning?

Answer to 7.15:

(a) The change from the battery being fully charged battery to being

totally discharged battery is a state function in both cases. The path is

quite different, but the end result is exactly the same, a dead battery. State

functions do not depend on the path taken to achieve the present state.

(b) In Case 1, much of the energy lost by the battery will be released as radiant

energy, light. Batteries do get warm as they are being used, also indicating that

there is some thermal energy transfer. In Case 2, there will be some thermal

energy transfer because of friction between the toy car and the roadbed on which

the car runs. Much of the energy lost will be changed to mechanical work done in

moving the toy car.

7.16. Consider a C-172 airplane (small private plane), a helicopter, a parachute jumper,

and a hawk, all presently at 1892 feet above the Clemson, South Carolina airport,

which is at an altitude of 892 feet. Each plans a landing at the airport. Comment

on the state functions and path functions for each of these airborne objects.

Answer to 7.16:

The state functions will be identical, as each object will make an altitude change

of 1000 feet to land safely on the ground. The path functions will be quite

different, however. The airplane and helicopter are obliged to follow certain

traffic patterns, with designated altitudes and descent rates into the airport. The

parachute jumper will likely be following a spiral pattern to land safely. The bird

will take whatever path it wants to, with little regard to the airplane, helicopter, or

parachute jumper. Interestingly enough, the bird also lands into the wind, just as

the other airborne objects plan to do!


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Section 7.5. System and Surroundings

7.17. Which ending(s) for the following sentence is(are) correct? Explain your

reasoning in each case. The enthalpy change for a system open to the atmosphere

is(a) dependent on the identity of the reactants and products

(b) zero

(c) negative

(d) qP

(e) positive

Answer to 7.17: (a) and (d)

7.18. Give “real life” examples of the following:

(a) an open system(b) a closed system

(c) an isolated system

(d) an exothermic process

(e) an endothermic process

Answer to 7.18:

(a) An open system refers to part of the universe being studied that can

exchange matter and energy with its surroundings. Some open system

examples include an open aquarium, a cup of coffee, and hot springs. Ourbody is also an open system since we exchange both energy (in the form

of heat) and matter (food, O2, CO2, H2O) with the surroundings.

(b) In a closed system, only energy is exchanged. An example of such

system is a cup with hot coffee that is tightly closed so no vapors may

escape but the cup is not isolated so heat energy may be exchanged with

surroundings.

(c) An isolated system does not exchange matter or energy. An example of

a nearly isolated system might be a properly stoppered Thermos® bottle.A stopper prevents water vapor from escaping, while the vacuum

construction keeps heat from being lost to the surroundings.

(d) An exothermic process is a process that releases heat to its

surroundings. Reactions of sodium metal with water, burning of a match

or solidification of molten metals are examples exothermic processes.


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(e) An endothermic process is a process that absorbs heat into its surroundings.

Melting ice or evaporation of rubbing alcohol are examples of endothermic

processes.

7.19.A properly stoppered Thermos bottle containing coffee is nearly an isolatedsystem. The stopper prevents water vapor from escaping, while the vacuum

construction keeps heat from being lost to the surroundings. How would you turn

this isolated system into an open system? A closed system?

Answer to 7.19:

An open system can exchange matter and energy between system and

surroundings. Opening the stopper would change Thermos bottle into open

system in which matter (vapors) and heat can be transferred. A closed system can

exchange energy with its surroundings, but no matter. If, for example, the stopper

of Thermos bottle is pierced with a piece of heavy gauge copper wire, heat

would leave the system by conduction along the copper wire. Removing the

vacuum construction would also convert the isolated system of Thermos bottle

into an open system in which heat can be exchanged through the walls of the

bottle.

7.20. The earth is sometimes described as a closed system, particularly when

considering regional or global environmental problems. For example, acid rain

may fall at great distance from the original site of emissions responsible for theobserved effect. Ozone depletion near the South Pole is linked to the emission of

fluorocarbons from uses in industrial nations. While this view does convey a

sense of the many connections that affect environmental issues, is it correct to

think of the earth as a closed system from the standpoint of thermodynamics?

Why or why not?

Answer to 7.20:

The earth is an open system because matter and energy is exchanged on a regular

basis with the surroundings. For example, molecules escape the earth’s surfaceregularly, placing gases that are responsible for acid rain into the atmosphere.

Meteorites fall to the earth from outer space. Thermal energy is transferred from

the earth to the atmosphere, and various types of electromagnetic energy are

transferred to the earth’s surface.


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7.21. This diagram shows an

experiment in which a hot silver

bar is added to a beaker of water

at room temperature.

(a) If the object of the

experiment is to find the specific

heat of the silver, define the

system and the surroundings.

(b) If you already know the specific heat of the silver, the same experimental set

up might be used to determine if there is heat loss through the walls of the glass

beaker or from the thermometer itself. Define the system and surroundings in that

experiment.

(c) Does this experimental diagram illustrate an open, closed, or isolated system?

Give the reasoning behind your choice.

Answer to 7.21:

(a) The silver bar and the water form the collection of atoms and

molecules under consideration, the system. Everything else, including the

beaker, the thermometer, the air above the water, and everything else in

the universe are surroundings. There will be thermal energy transfer from

the warmer silver bar to the cooler water molecules until equilibrium is

reached. At that point, the heat released by the silver bar equals the heat

gained by the water. It is necessary to know the mass of the silver bar, the

mass of water present, and the specific heat of water.

(b) The silver bar, the water, the beaker, and the thermometer are all part

of the system in this experiment. Everything else forms the surroundings.

(c) This is an open system because matter and energy can be exchanged with the

surroundings. For example, water molecules can evaporate from or condense onto

the surface of the water. Heat may be exchanged between the system and the

surroundings. Therefore, this cannot be a closed system because matter can be

exchanged. It is not an isolated system as there is no attempt to control thermal

exchanges between the system and the surroundings.


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Section 7.6. Calorimetry and Introduction to Enthalpy

7.22. In a calorimetric measurement, why might it be important to know the heat

capacity of the calorimeter? The calorimeter heat capacity is the amount of

thermal energy the calorimeter (container, thermometer,etc

.) transfers per degreetemperature change.

Answer to 7.22:

To accurately measure the thermal energy change for reaction, we need to account

for heat that goes to heating (in the case of exothermic reaction) or cooling (in the

case of endothermic reaction) the calorimeter itself. The heat lost or gained to the

calorimeter must be experimentally determined for each calorimeter to assure the

accuracy of calorimetric measurements.

7.23. Choose the best response to complete the final sentence. Explain why you makethis choice and what is wrong with each of the others. The equation for

decomposition of gaseous ammonia, NH3(g), is:

NH3(g) → 1/2N2(g) + 3/2 H2(g) ∆ H = -45.9 kJ

This reaction equation and enthalpy change indicate that the formation of gaseous

ammonia

(a) evolves 45.9 kJ for each mole of ammonia formed.

(b) evolves 23 kJ for each mole of nitrogen used.

(c) absorbs 45.9 kJ for each mole of ammonia formed.(d) absorbs 23 kJ for each mole of nitrogen used.

(e) is an exothermic process.

Answer to 7.23: (c)

7.24. A 10.0 g sample of an unknown metal was heated to

100.0 °C and then added to 20.0 g of water at 23.0 °C in

an insulated calorimeter. At thermal equilibrium, the

temperature of the water in the calorimeter was 25.0 °C.

Which of the metals in this list is most likely to be the

unknown?

Answer to 7.24:

When the hot metal is placed in the cooler water, the heat lost by the metal

and the heat gained by the water sum to zero.


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Specific Heat Data

H2O(l) 4.184 J·g–1·°C–1

Au(s) 0.13 J·g–1·°C–1

Ag(s) 0.22 J·g–1·°C–1

Al(s) 0.90 J·g–1·°C–1


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heat lost + heat gained = 0

heat lost = –heat gained

mmetal × smetal × (Tf, metal–Ti, metal) = –m H2O × s H2O × (Tf, H2O–TI, H2O)

10 g × smetal × (25.0 °C – 100.0 °C) = –20 g water × 4.184 J

g ⋅°Χ × (25.0 °C – 23.0

°C)

specific heat metal = 0.22 J·g-1·°C-1 The metal is likely to be silver.

7.25. A student received 55.0 g of unknown metal from his laboratory instructor. To

identify this metal, he decided to calculate its specific heat. He heated the metal

sample to a temperature of 98.6 °C in a water bath. Then he transferred it slowly

to a calorimeter containing 100. mL of water at 24.6 °C. The maximum

temperature reached by the calorimeter system was 25.8 °C. His lab instructor,

who was watching the experiment, explained to the student that he hadintroduced a large amount of error and asked him to repeat the experiment with

the correct procedure. In the second trial, the temperature of the water bath was

98.4 °C and the initial temperature of water in the calorimeter was 24.3 °C. The

maximum temperature after the sample was immersed in the calorimeter was 26.5

°C. What error had the student made? Explain how you know.

Answer to 7.25:

The student transferred the hot material into the calorimeter too slowly. The heat

lost from metal was transferred to the air outside the calorimeter and it was not

equal to the heat gained by the water in the calorimeter. Therefore, the student

measured a lower final temperature of the water for the first trial.

7.26. A student mixed 100. mL of 0.5M HCl and 100. mL of 0.5M of NaOH in a

Styrofoam® cup calorimeter. The temperature of the solution increased from

19.0 °C to 22.2 °C. Is this process exothermic or endothermic? Assuming that the

calorimeter absorbs only a negligible quantity of heat, that the density of the

solution is 1.0 g·mL–1, and that its specific heat is 4.18 J·g–1·°C–1, calculate the

molar enthalpy change for the reaction:HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

Answer to 7.26:

Total volume of solution is 200 mL . The mass of the solution is: (200

mL) x 1g/mL = 200 g

The temperature change is 22.2oC – 19.0oC = 3.2oC.


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The temperature increased. This means that thermal energy is released

from the system and the reaction is exothermic.

The heat gained by water is given by the product of the mass, specific

heat, and temperature change.

qp = (200 g) x (4.18 J/goC) x (3.2oC) = 2700 J =2.7 kJ

The heat gained by the system (reacting chemicals) is q= -2.7kJ.

To find the heat evolved per mole of reactants (enthalpy change) we

have to calculate the number of moles.

(0.1L) x (0.5 mol/L) = 0.050 mole

∆ H = -2.7kJ / 0.05 mole = -54kJ/mole

7.27. Instant cold packs, such as those used to treat athletic injuries, contain ammoniumnitrate and a separate pouch of water. When the pack is activated by squeezing to

break the water pouch, the ammonium nitrate dissolves in water. This is an

endothermic reaction and the change in enthalpy is 25.7 kJ·mol –1 of ammonium

nitrate dissolved.

(a) Why does this endothermic reaction produce a cold sensation on your skin?

(b) The cold pack contains 125 g of water to dissolve 50.0 g of ammonium

nitrate. What will be the final temperature of the activated cold pack, if the initial

room temperature is 25 °C? Assume that the specific heat of the solution is thesame as that for water, 4.184 J·g-1·°C-1.

(c) What other assumptions do you make in carrying out the calculation in part

(b)?

(d) The final temperature is about 7 °C. Does this result justify the assumptions

in parts (b) and (c)? If not, explain which assumption(s) might not be valid.

Answer to 7.27:

(a) The solution process is endothermic and is drawing heat from your

bruised skin to enable the ammonium nitrate to dissolve.(b) If the cold pack contains ample water to dissolve 50.0 g of ammonium

nitrate,

50.0g NH 4NO3 ×1 µολ ΝΗ 4ΝΟ380.0 γ ΝΗ 4 ΝΟ3

×25.7 κϑ

µολ ΝΗ 4ΝΟ3×

103 ϑ

κϑ= 1.61×10

4 ϑ is


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the heat absorbed by the dissolving ammonium nitrate. This is equal to

the heat absorbed by the water.

−1.61 ×10 4 J = 175 g H 2O ×4.184 J

g ⋅°C× (X °C − 25 °C)

X = 3.

1 °Χ

(c) It is assumed that there is no heat loss or gain between the heat pack and the

surroundings.

Section 7.7. Bond Enthalpies

7.28. Ethanol, by itself or in blends, is an important alternative fuel for gasoline-

powered engines. The molecular equation for the complete combustion of ethanol

is:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

(a) Rewrite the balanced equation using full Lewis structures for each reactant

and product molecule.

(b) Use your model set and build a model of each reactant and product molecule

in this reaction.

(c) Refer to the Lewis structures, the models, and Table 7.3 to determine the

total enthalpy input that will be required to break all the bonds in the reactants.

(d) Refer to the Lewis structures, the models, and Table 7.3 to determine the

total enthalpy that will be released in forming all bonds in the products.

(e) What is the net enthalpy change in this combustion reaction? Is your result

likely to be an overestimate or underestimateof the experimental ∆ H oreaction?

Explain your reasoning.

(f) Ethanol is also a fuel when consumed by humans. Will your result in part (e)

also give an estimate for ethanol’s fuel value in our bodies? Why or why not?

Answer to 7.28:

(a)

Answer 7.1.

C

R

H

C

H

H

O +H H O O3 O C O2 + 3 O H

H

Answer 7.2. (b) Use your model set and build a model of each reactant

and product molecule in this reaction.


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Answer 7.3. (Could insert photo of models here)

Answer 7.4. (c) ∑ BH reactants = (5 × BH C–H + 1 × BH C-C + 1 × BH C-O + 1

× BH O-H) + (3 × BH O=O)

= [(5 mol × 414 kJ·mol-1) + (1 mol × 347 kJ·mol-1) + (1 mol × 351

kJ·mol-1

) + (1 mol × 460 kJ·mol-1

)] + [(3 mol × 499 kJ·mol-1

)= 4725 kJThis is the energy required to break all bonds in reactants. The sign is

positive showing bond breaking is an endothermic process.

(d) ∑ BH products = (2 × 2 × BH C=O + 3 × 2 × BH O-H)

= [(4 mol × –745 kJ·mol-1) + (6 mol × –460 kJ·mol-1) = –5740 kJ This is

the energy released in forming all bonds in products. The sign is negative

showing that bond formation is an exothermic process.

(e) ∆ H = ∑ BH reactants + ∑ BH products = +4725 kJ + –5740 kJ = –1015 kJ The

combustion of alcohol is an exothermic process overall, just as is expectedif the substance is being used as a fuel. The experimental reaction enthalpy

will involve a reactant (ethanol) and product (water) in their liquid states,

but the calculations based on bond enthalpies assume all species are gases.

Since this is an exothermic reaction, the ethanol is at a higher energy than

the products. Converting gaseous ethanol to liquid will lower its energy,

so the effect would be to reduce the experimental relative to calculated

energy. Converting gaseous water to liquid will lower its energy and theeffect would be to widen the energy gap between the reactants and

products and thus increase the experimental relative to calculated energy.

Which effect would predominate? Three moles of water are liquefied, but

only one mole of ethanol. The molar enthalpy of vaporization (or

condensation) of ethanol and water are about the same (Table 1.2 in

Chapter 1), so the water effect would be larger and the calculated energy

would likely be an underestimate. (The value calculated from standardenthalpies of formation – Section 7.8 – is about -1368 kJ·mol–1, which, as

predicted, is somewhat larger than the value, -1231 kJ·mol–1, from bond

enthalpies.)


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(f) The calculation of the enthalpy of combustion was done using average

bond enthalpies for gaseous compounds at 25°C and there is some error in

using such values for compounds in other states. There surely would also

be some differences if combustion through metabolic processes were to

take place. Still, values for food energy are determined through

combustion reactions carried out in typical calorimetric experiments.

7.29. Ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3, are isomers. Use average

bond enthalpies to determine which of the isomers is the more stable. By how

much? Clearly explain your reasoning and the calculations you do to get your

result.

Answer to 7.29:

The more stable isomer will require greater energy to dissociate it into

atoms. We can use bond enthalpies to calculate the enthalpy change for

atomization of each isomer and compare the results.

ethanol dimethyl ether

bond # bonds BH, kJ # bonds BH, kJ

C-H 5 2070 6 2484

C-C 1 347

O-H 1 460

C-O 1 351 2 702

Total 3228 Total 3186

7.30. Ethanol requires more enthalpy to atomize to the same collection of atoms, so it is

the more stable of these two isomers by about 42 kJ (per mole of

isomer).Consider this reaction in which two amino acids join to form a peptide

bond, releasing a water molecule:

C C

H

N

R

H

H

O

O H

C C

H

N

R'

H

H

O

O H

C C

H

N

R

H

H

O

N C

H

C

O

O HR'

H

O

H

H

Use average bond enthalpies to estimate the enthalpy change for this reaction.

Answer to 7.30:

∆ H = ∑ BH reactants + ∑ BH products= (1 × BH C–O + 1 × BH N–H) + (1 × BH C–N + 1

× BH O–H)= [(1 mol × 347 kJ·mol-1) + (1 mol × 393 kJ·mol-1)] +

[(1 mol × 276 kJ·mol-1) + (1 mol × 460 kJ·mol-1)] = 740 kJ + 736

kJ = +4 kJ


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7.31. The glucose molecule can exist in a number of different forms, including the

open-chain and cyclic forms represented below. By convention, the carbons in thechain are numbered 1 through 6 beginning at the top, as shown. Carbon-1 in the

cyclic form is the one at the far right of the structure. To make the cyclic form, the

oxygen bonded to carbon-5 in the linear form becomes the ring oxygen in the

cyclic form where it is bonded to both carbon-1 and carbon-5. Which is the more

stable, the open-chain or the cyclic structure? How much more stable? Show your

reasoning and calculations clearly.

C

C

C

C

C

C

H

H

OHH

HHO

OHH

H

OHH

OH

1

2

3

45

6

O

C

CC

C

OC H

H

H

O

H

O

H

O

COH

HH

H

HO

H

H

open chain cyclicAnswer to 7.31:All the bonds in the two forms are the same, except those involved in

making the cyclic form from the open-chain form. The change involves

breaking a C=O on carbon-1 and making two C-O bonds in its place.

An O-H bond also has to be broken, but another is made, so they cancel out. Thus,

the difference is 745 kJ for C=O compared to 702 kJ for 2 C-O. By this

calculation, the open-chain form would be more stable (require more energy to

atomize) by about 43 kJ than the cyclic form. In nature, almost all glucose


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molecules exist in cyclic forms. There must be another stabilizing influence (or

bond enthalpy calculations for glucose, which is a solid, are not accurate enough

to give us the actual enthalpy difference).

7.32.Two isomers with the molecular formula CH2N2 are:

C N

H

H

NC

N

N

H

H

diazomethane diazirine

Diazomethane and diazirine are both gases. Under appropriate conditions, each

can react to give ethene and nitrogen:

2CH2N2(g) → H2C=CH2(g) + 2N2(g)

(a) Use average bond enthalpies to calculate ∆ H °reaction for diazomethane and for

diazirine undergoing this reaction. Are the reactions exothermic or endothermic?(b) Construct enthalpy-level diagrams (like Figure 7.11) for the two reactions.

Which isomer is the more stable? Are there any factors that might complicate use

of bond enthalpies for these molecules?

(c) Does our molecular bonding model help explain the relative stability?

Explain why or why not.

Answer to 7.32:

(a) Note that, for both reactions, we can ignore the C-H bonds, since there are

four in reactants and four in the products. We have to focus on the carbon-nitrogen, and nitrogen-nitrogen bonds.

For the reaction of two moles of diazomethane, we have:

∆ H rxn = – BH (C=C) –2 BH (N≡N) + 2 BH (C=N) + 2 BH (N=N)

∆ H rxn = – 620 – 2·(941.4) + 2·(615) + 2·(418) = –437 kJ

For the reaction of two moles of diazirine, we have:

∆ H rxn = – BH (C=C) –2 BH (N≡N) + 4 BH (C–N) + 2 BH (N=N)

∆ H rxn = – 620 – 2·(941.4) + 4·(276) + 2·(418) = –563 kJ

These reactions are highly exothermic with the predominant contribution in bothcases coming from the great stability of the triple-bonded nitrogen product. Note

that that only difference between these two calculations is the breaking of

different kinds of carbon-nitrogen bonds. For diazomethane, two C=N bonds are

broken and, for diazirine, four C–N bonds are broken. Recall that the enthalpy

change required to break one C=N bond is greater than twice the enthalpy change


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to break two C–N bonds. The difference, 63 kJ. For two moles of diazomethane

and diazirine, the difference in reaction enthalpies is 126 kJ (= 2·63 kJ), reflecting

the difference in carbon-nitrogen bonds broken.

(b)

Diazomethane is the more stable molecule, relative to diazirine. Complicatingfactors could be the strain of making the three-membered ring (which distorts the

usual bonding angles for carbon and nitrogen) in diazirine and/or the formal

separation of charge in the diazomethane molecule that could make it less stable

than a comparable molecule without formal charge.

(c) You might argue that the strain energy in diazirine is responsible for its higher

energy (63 kJ·mol–1 higher, based on these bond enthalpy calculations), but the

argument is not wholly satisfactory. To complicate matters, we find

experimentally that diazirine can be kept as a gas in the dark at room temperatureindefinitely, whereas samples of gaseous diazomethane react as shown here and

on the walls of their containers within a few hours. Based on exothermicity, the

reactions shown here seem very favorable, although we know that exothermicity

is not a good criterion for deciding the direction of a reaction. Diazomethane does

react, but some factor slows the diazirine reaction so much that it is undetectable.


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7.33. The table at the right gives the molecular

formulas and enthalpies of combustion for

three common sugars found in living

organisms. What conclusions can you drawfrom these data about the bonding in these

molecules? Explain your response.

Answer to 7.33:

As you see, these molecules are isomers, and their complete combustion will give

identical products in each case: 6 molecules of CO2 and 6 molecules of H2O for

each molecule of sugar oxidized. The identical values of their enthalpies of

combustion (to within 0.3%) indicates that the bonding must be essentially

identical in these isomers. That is, there must be the same number of C–C, C–H,

C–O, C=O, and O–H bonds in all three molecules. This is true if you compare

their open-chain forms with one another or their cyclic forms with one another (in

nature, the cyclic forms vastly predominate).

7.34. Disaccharides are sugars formed by the combination of two simpler sugars. A

maltose molecule, for example, is a combination of two glucose molecules:

C

C

C

C

OC H

H

H

O

H

O

H

O

COH

H H H

HO

H

H

C

C

C

C

OC H

H

H

O

H

O

H

O

COH

H H H

HO

H

H

C

CC

C

OC H

H

H

O

H

O

H

O

C

HH

H

HO

H

H

C

CC

C

O

C H

H

H

O

H

O

C

OH

HH

HO

H

H

O

H

H2

O

(a) Estimate ∆ H °reaction, using bond enthalpies. Is it necessary to do any numerical

calculations to make this estimate? Explain why or why not.

(b) Use your result from part (a) and the data from the Problem 7.33 to estimate

the enthalpy of combustion of maltose. The experimental value is –5644 kJ·mol –1.

How well does your estimate compare to the experimental value? How do you

explain the agreement or lack of agreement?

Answer to 7.34:

(a) In this reaction a C–O bond is broken in one sugar and an O–H bond in

the other. Then to make the products, a C–O bond is made between the

two sugars and an O–H bond is made to produce the water (from the OH

that had left a sugar). Exactly the same number and kind of bonds are

broken and made. Bond enthalpy calculations (which we do not have to


22

Sugar Formula∆ H

combustion

kJ ⋅mol–1

fructose C6H12O6 –2812

galactose C6H12O6 –2803

glucose C6H12O6 –2803


23/53

do, if there is no change in the kind and number of bonds) give an

estimated ∆ H reaction = 0 kJ for this reaction.

(b) One way to estimate the enthalpy of combustion of maltose is to break

the reaction into two reactions: hydrolysis of maltose to give two glucose

and then combustion of the glucose:

maltose + H2O → 2 glucose ∆ H rxn = 0 kJ

2 glucose + 12 O2 → 12 CO2 + 12 H2O ∆ H rxn = – 5606 kJ (from 7.17)

The sum of these two reactions is:

maltose + 12 O2 → 12 CO2 + 11 H2O ∆ H rxn = – 5606 kJ

This is the combustion reaction for maltose; we predict ∆ H rxn = – 5606 kJ to be

compared with the experimental value of –5644 kJ. The difference is small (only

about 0.6%), but the combustion enthalpies are experimental values that are good

to at least four significant figures. This means that the estimated value of 0 kJ for

the formation of maltose (or the reverse, its hydrolysis) is incorrect. The actual

value for the hydrolysis above must be about –36 kJ, in order to make the sum of

the two reaction enthalpies above equal to the measured enthalpy of combustion

of maltose. We know that bond enthalpies are averages that can be off by a few

percent for any particular bond in a molecule, so we shouldn’t be surprised that

this small discrepancy occurs. It is a warning, however, not to trust bond

enthalpies when the quantity of interest is the difference between two large values

-- small errors in the large values can lead to large percentage discrepancies in

their difference.

7.35. The reaction between hydrazine and hydrogen peroxide in rocket engines can be

represented as:

H2NNH2(g) + 2HOOH(g) → N2(g) + 4H2O(g)

(a) Use average bond enthalpies to estimate the standard enthalpy change (in kJ

per mole of hydrazine) for the reaction.(b) Compare the energy released in the hydrazine reaction to the energy that

would be obtained if two moles of ammonia were oxidized by hydrogen peroxide.

2NH3(g) + 3HOOH(g) → N2(g) + 6H2O(g)

(c) Which provides the most energy per gram of fuel, hydrazine or ammonia?

Answer to 7.35:


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(a) Let’s cancel the H–O bonds before going through all the arithmetic.

Four are broken on the left and eight are made on the right, so we will

include four in the products, but ignore them in the reactants. The ∆ H rxn

for this reaction is given by:

∆ H rxn = – BH (N≡N) –4 BH (H–O) + 4 BH (H–N) + BH (N–N) + 2 BH (O–O)

∆ H rxn = – 941 – 1840 +1572 + 193 + 284 = – 732 kJ

(b) In this case, canceling the H–O bonds leaves a net of six in the

products:

∆ H rxn = – BH (N≡N) –6 BH (H–O) + 6 BH (H–N) + 3 BH (O–O)

∆ H rxn = – 941 – 2760 + 2358 + 426 = –917 kJ

Two moles of ammonia reacting release about 25% more enthalpy than a

mole of hydrazine.

(c) A mole of hydrazine is 32 g of hydrazine, so the enthalpy change from

hydrazine is – 22.88 kJ·g–1 ((= –732 kJ)/(32 g)). The comparable value for

ammonia is –26.97 kJ·g–1 ((= –917 kJ)/(34 g)).

Since more enthalpy is released by the ammonia reaction, why isn’t ammonia

used, instead of hydrazine as the fuel? There are quite practical reasons for using

hydrazine. Probably the most important, from an engineering point of view, is that

hydrazine is a liquid at room temperature, Table 7.4, and can be easily stored and

pumped around in the rocket. Ammonia, on the other hand, is a gas at roomtemperature and has to be refrigerated and/or kept under high pressure to store it

as a liquid. This just increases the complexity of the rocket. Consider also the

oxidant, hydrogen peroxide. About one-and-one-half times as much peroxide is

required for the oxidation of ammonia as for an equal mass of hydrazine.

Therefore, the amount of fuel and oxidant together is less for hydrazine. The same

size rocket can carry a somewhat greater payload with the hydrazine fuel because

less mass of fuel and oxidant are needed. Work out for yourself the enthalpy

change for reaction of one gram of the stoichiometric mixture of fuel and oxidantfor each fuel.

7.36. WEB Chap 7, Sect 7.7.2-3. On Web Companion page 7.7.2 you select the

bonds that are broken in the reactants and made in the products.

(a) Use the data on page 7.7.3 to calculate the enthalpy change for this process.


24

WEB


25/53

(b) How does your result in part (a) compare with the analysis on page 7.7.3

where all reactant bonds are broken and all product bonds formed?

(c) Explain why the comparison in part (b) comes out the way it does.

Answer to 7.36:

(a) The two bonds that have to be broken in the reactants and the three that haveto be made in the product are indicated here:

C

H

H

H C H

O

H

H

C

H

H

H C H

O

H

H

Breaking the H–O and C=C bonds requires 460 + 620 = 1080 kJ·mol–1. Making

the C–H, C–O, and C–C bonds releases energy, so the enthalpy change is

negative, – (414 + 351 + 347) = –1112 kJ·mol–1. The net change in enthalpy is the

sum of these two values, 1080 + (–1112) = –32 kJ·mol–1.

(b) On Web Companion page 7.7.3, you find that breaking all the bonds in the

reacatants requires 3196 kJ·mol–1 and making all bonds in the product releases –

3228 kJ·mol–1. The net change for the reaction is –32 kJ·mol–1. The results for the

two methods are identical.

(c) Finding that the results of the two methods are identical is expected. When all

the bonds in the reactants are broken (atomizing the molecules), we are breaking

several kinds of bonds, several C–H bonds, for example, that will simply be

remade in the product. Our assumption in using average bond enthalpies is that

the bond enthalpy for a given kind of bond is the same in all molecules. Thus

when the bonds are remade in the product, exactly the same amount of enthalpy isreleased as was required to break the bonds in the first place -- the breaking and

making simply cancel out. The only bond breaking and making that contribute to

a net difference for the reaction are those that are different in the reactants and

product, which is what we focused on in part (a). You can see this argument

graphically on Web Companion page 7.7.3, if you first break the two bonds

selected on page 7.7.2 and then break all the others to atomize the reactants. Now

when you make the product bonds, first make all those that are the same in the

product as in the reactants and then, finally, make the three new bonds. Comparethe lengths of the bars on the graph for the breaking and making the same set of

bonds.

7.37. Carbon–hydrogen bond-dissociation enthalpies are shown for two different

hydrogens in the reactions of propene and 2-butene shown here. The enthalpies

for the two different bond dissociations are quite different. The dissociation of a


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carbon–hydrogen bond on a carbon that is next to a doubly bonded carbon,

reactions (1) and (3), requires less energy.

CH

H

H

C C

H

H

H

CH

H

H

C C

H

C

H

H

H

H

CHH

C CH

H

H

CH

H

H

C CH

H

CH

H

C C

H

C

H

H

H

H

CHH

H

C C

H

C

H

HH H

H

H

H

dissociation enthalpy

kJ mol–1

361

430

358

430

(1)

(2)

(3)

(4)

(a) Are the propyl and butyl free radicals formed in reactions (1) and (3),

respectively, more stable (lower enthalpy) or less stable (higher enthalpy) than

those formed in reactions (2) and (4)? Clearly explain the reasoning for your

response, using enthalpy level diagrams, if these aid your explanation.

(b) The unpaired electron in the three-carbon and four-carbon free radicals

formed in reactions (1) and (3) is often shown as participating in delocalized π

bonding (see Chapter 5, Section 5.7), with the two π electrons from the adjacent

double bond. Write Lewis structures supporting this model. Do the bond-

dissociation enthalpies and your interpretation in part (a) support this model?

How would delocalization of the unpaired electron orbital affect the energy of the

radical? Explain your reasoning clearly.

Answer to 7.37:

(a) The enthalpy level diagram at the right represents the enthalpies of the

reactant and the two sets of products from reactions (1) and (2).


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It takes less enthalpy to produce the

products from reaction (1). Thus, the free radicals formed in reaction (1)

{and (3)} are more stable (by 69 kJ) than those formed in reaction (2)

{and (4)}.

(b) The “equivalent” Lewis structures for the three- and four-carbon free radicals

formed in reactions (1) and (3) are:

CH

H

C C

H

H

HC H

H

CC

H

H

HCH

H

C C

H

C

H

H

H

H CH

H

C C

H

C

H

H

H

H

The structures for the three-carbon radical are identical in energy, so we know

from our discussion in Chapter 5, Section 5.7, that the electrons involved are

delocalized and of lower energy than they would be if the two pi double-bond

electrons and single free-radical electron did not interact. The two four-carbon

structures are not quite identical, but the sigma bonding to the carbon atoms with

the free radical electron is very similar in the two structures (to two hydrogen

atoms in the first and to a hydrogen atom and carbon atom in the second), so we

can treat the structures as energy-equivalent and again see that there is electron

delocalization to lower the energy of this free radical

Section 7.8. Standard Enthalpies of Formation

7.38. You usually see sulfur in the form of a light yellow powder, the rhombic

crystalline allotrope of sulfur. If you melt this solid (m. p. 112.8 °C), pour the

liquid into a filter paper cone, and then unfold the cone as the sample cools, you

find that the liquid has solidified in dark yellow needles. This is the triclinic

crystalline allotrope of sulfur. After several days, the needles become covered by


27


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a light yellow layer. Which allotrope of sulfur is probably its most stable form

under standard conditions? Explain your answer.

Answer to 7.38:

The rhombic allotrope, the light yellow powder, is the more stable at room

temperature. At higher temperatures, near the melting point, the triclinic allotrope

must be more stable, since it is the form that crystallizes from the molten sulfur.

Upon standing at room temperature, the rhombic allotrope forms slowly on the

surface of the triclinic needles, which indicates a slow transformation to the more

stable form. If we wait long enough, all the needles change back to the light

yellow powder.

7.39. Urea (Latin urina = urine), H2NC(O)NH2, is a water-soluble compound made by

many organisms, including humans, to eliminate nitrogen. Use standard

enthalpies of formation to find the enthalpy for the reaction producing urea from

ammonia and carbon dioxide.

2NH3(g) + CO2(g) → H2NC(O)NH2(s) + H2O(l)

Show all your work so your procedure is clear.

Answer to 7.39:

The standard enthalpies of formation for ammonia, carbon dioxide, urea,

and water are – 46.1, –393.5, –333.0, and –285.8 kJ·mol–1, respectively.

The standard enthalpy change for the reaction is (where the stoichiometric

coefficients -- units of mol -- are not shown, if they are unity):

∆ H orxn = ∆ H of (H2O(l)) + ∆ H

of (H2NC(O)NH2(s)) – 2∆ H

of (NH3(g)) – ∆ H

of (CO2(g))

∆ H orxn = (–285.8 kJ) + (–333.0 kJ) – (– 92.2 kJ) – (– 393.5 kJ) = –133.1 kJ

7.40. In Problem 7.39, you used enthalpies of formation to calculate the enthalpy for

making urea from ammonia and carbon dioxide. How could you estimate the

enthalpy of vaporization of urea? Explain the procedure you would use and then

carry it out. Hint : What other way could you estimate the enthalpy of the reaction

forming urea? How do the two ways differ?Answer to 7.40:


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Recall that bond enthalpy values are

applicable to gas phase molecules. We

could use bond enthalpies to calculate the

enthalpy change for the reaction in 7.25

with the urea and water as gases and

compare the result with the enthalpy

change calculated in 7.25 for the urea as a

solid and water as a liquid. The difference

is an estimate of the enthalpy of

sublimation of urea plus the enthalpy of

vaporization of water. (The problem is

misstated when it asks for the enthalpy of

vaporization of urea. You can get the

enthalpy of sublimation, but have no way

to estimate the changes from solid to

liquid to gas.) For the gas phase reaction

we have:

∆ H orxn = –2 BH (H–O) –4 BH (H–N) – 2 BH (C–N) – BH (C=O) +

6 BH (H–N) +

2 BH (C=O in CO2)

∆ H orxn = – 920 – 1572 – 552 – 745 + 2358 + 1598 = + 167 kJ

Enthalpy is required to make gaseous urea and water from ammonia and carbon

dioxide. The enthalpy change (decrease -- exothermic change) going from the

gases to solid urea and liquid water is -133 – (167) = – 300 kJ. The enthalpy of

vaporization of water is 44 kJ·mol–1, so the condensation of water accounts for –

44 kJ. This leaves – 256 kJ as the enthalpy released when gaseous urea condenses

to solid urea. Our estimate for the enthalpy of sublimation of urea is about 256

kJ·mol–1.

7.41. During fermentation of fruit and grains, glucose is converted to ethanol and

carbon dioxide according to this reaction:

C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

(a) Using the data in Appendix XX, calculate ∆ H °reaction.

(b) Is this reaction exothermic or endothermic?


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(c) Which has higher enthalpy, the reactant or the products?

(d) Calculate ∆ H for the formation of 5.0 g of C2H5OH(l).

(e) What quantity of heat is released when 95.0 g of C2H5OH(l) is formed at

constant pressure?

Answer to 7.41: (a) ∆ H °reaction = -68.0 kJ•mol-1 (b) exothermic (c) reactants

(d) -3.7 kJ (e) -70.2 kJ

7.42. Consider this gas phase reaction in which methanoic (formic) acid reacts with

ammonia. The product formamide contains a peptide bond and there is the release

of a water molecule:

H C

O

O H

HN

H

H

H C

O

N H

H

O

H

H

(a) Use average bond enthalpies to estimate the enthalpy change for this

reaction. How does this ∆ H ° compare to the ∆ H ° you calculated in problem 7.29?

Would you have expected this result? Why or why not?

(b) Use standard enthalpies of formation to calculate the change in enthalpy for

this reaction, assuming reactants and products are gases in their standard states.

The enthalpies of formation of gaseous methanoic acid and formamide are –379

and –186 kJ·mol–1, respectively.

(c) Compare the results of your calculations in parts (a) and (b). Offer some

reasons to help explain your comparison.

Answer to 7.42:

(a) ∆ H = ∑ BH reactants + ∑ BH products= (1 × BH C–O + 1 × BH N–H) + (1 × BH C–N +

1 × BH O–H)= [(1 mol × 347 kJ·mol-1) + (1 mol × 393 kJ·mol-1)] + [(1 mol

× 276 kJ·mol-1) + (1 mol × 460 kJ·mol-1)] = 740 kJ + 736 kJ= +4 kJ

Note that the calculation using average bond enthalpies is the same as in

the previous problem, for exactly the same bonds are broken and formed.

(b) Calculation using standard enthalpies of formation.

∆ H reaction = Σn p∆ H 0

f (products) – Σnr ∆ H 0

f (reactants)

= [(1 mol) × –254 kJ·mol-1) + (1 mol) × –285.8 kJ·mol-1)] –

[(1 mol) × –363 kJ·mol-1) + (1 mol) × –46.3 kJ·mol-1)]= –130.5 kJ


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(c) The two calculations are not very close. Average bond enthalpies are based on

gaseous compounds at 25 °C. Standard enthalpies of formation are based on the

assumption that reactants and products are in their standard states at 25 °C.

Neither set of assumptions can be expected to hold true for this reaction, which

likely takes place in aqueous solution. The bond enthalpies within the water

molecule are given for the gaseous state, but the standard state for water at 25 °C

is liquid. Even correcting for the Enthalpy of Vaporization for water does not

yield that close a result, because it still would be necessary to correct for the

Enthalpy of Solution values for the other reactants and products.

7.43. The combustion of ammonia is represented by this equation:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Experimentally, we find ∆ H °reaction = – 905 kJ for the reaction as written.

(a) Use the standard enthalpies of formation for NO(g) and H2O(g) from

Appendix XX and the standard enthalpy change for the reaction to calculate the

standard enthalpy of formation for NH3(g).

(b) How does your result in part (a) compare with the standard enthalpy of

formation for NH3(g) in Appendix XX? Explain why they are the same (or

different).

Answer to 7.43:

(a) ∆Hreaction = Σnp∆H0f (products) – Σnr ∆H

0f (reactants)

–905 kJ = [(4 mol × 90.4 kJ·mol-1) + 6((1 mol × –241.8 kJ·mol-1)] –

[4(∆ H 0f (NH 3) + 5(0)]

–905 kJ = [361.6 kJ – 1450.8 kJ] – 4(∆ H 0f ( NH 3)

∆ H 0f ( NH 3) = –46.0 kJ·mol-1

(b) The table value is –46.3 kJ·mol-1. This is excellent agreement and points up

the fact that the table values are often determined by means of values for

combustion reactions which are relatively straightforward to run.

7.44. In the process known as coal gasification, coal can be reacted with steam and

oxygen to produce a mixture of hydrogen, carbon monoxide, and methane gases.

These gases are desirable as fuels and they can also serve as the starting material

for the synthesis of other organic substances such as methanol, used for the


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production of synthetic fibers and plastics. Consider these three reactions that can

take place in the process of coal gasification.

Reaction 1: C(s) + H2O(g) → H2(g) + CO(g)

Reaction 2: C(s) + 1 / 2O2(g) → CO(g)

Reaction 3: C(s) + 2H2O(g) → CH4(g) + O2(g)

Use standard enthalpies of formation to find the enthalpies of reaction for each of

these three reactions. If it is possible to control the reaction conditions to favor

one or more of these reactions, which is the most energetically favorable? the

least energetically favorable? Explain your reasoning.

Answer to 7.44:

(a) Reaction 1: ∆ H orxn = +373.1 kJ

Reaction 2: ∆ H orxn = –110.5 kJ

Reaction 3: ∆ H orxn = –393.5 kJ

(b) The most exothermic reaction is Reaction 3. This does not imply that chemists

should work to control the reaction conditions to maximize this reaction. The

main goal of coal gasification is to produce fuels and starting materials for

synthesis. CO and H2 are the desired products, but the reaction to produce carbon

dioxide takes place at the same time.

Section 7.9. Harnessing Energy in Living Systems

7.45. What is a coupled reaction? What is its importance in biological reactions?

Answer to 7.45:

In a coupled reaction, the reaction providing the energy does not proceed unless

the reaction requiring the energy occurs. For example, the exothermic oxidation

of glucose is coupled with the endothermic formation of adenosine diphosphate,

ADP3-, from adenosine triphosphate, ATP4-. The reverse reaction, the hydrolysis

of ATP4- to ATP3- is exothermic and is coupled to other energy-requiring

biological functions such as locomotion, information processing, and synthesizingnew biological molecules.

7.46. What is the role of ATP in biological reactions?

Answer to 7.46:

The function of ATP is to store energy when energy is not needed for biological

reactions and to release it instantly when biological reactions require it.


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7.47. Coupled reactions take place in living systems, but coupled processes have many

applications in industry as well. For example, waste heat from a power plant can

be captured and used to do work, such as desalting seawater.

(a) Other than the availability of seawater, what do you think might limit the

usefulness of this example of a coupled process?

(b) Can you think of any other examples of energetically coupled processes from

your personal experience or other courses you have taken? Briefly describe such a

process. You might find it useful to research this topic using web-based resources.

Answer to 7.47:

(a) There are limitations on the overall efficiency of this coupled process.

One of the major limiting factors is using waste heat from a power plant

for desalting seawater is that the heat content of the waste heat is relatively

low. It is not great enough to boil water, so a process of flash distillation is

used. This coupled process is not widely used in the world, although solar

distillation processes are more common.

(b) Answers will vary depending on the background of the students. They may

present examples from engineering, other biological applications, or from ecology

courses. Note: This question could be assigned as a web-based research question.

7.48. Many ATP4--coupled reactions require that Mg2+ be present as well. What do you

think is the function of Mg2+ in these reactions? Hint : See Chapter 6, Section 6.6.

Answer to 7.48: Mg2+ can complex with the negatively charged ATP ions,

providing a three dimensional scaffold that allow these coupled reactions to occur.

7.49. The aerobic reaction sequence of glycolysis involves the complete oxidation of

pyruvic acid, CH3C(O)C(O)OH (= C3H4O3). This reaction is coupled to the

formation of ATP4-, according to this overall reaction:

2C3H4O3(l) + 5O2(g) + 30ADP3-

(aq) + 30HOPO32-

(aq) + 30H+(aq) →

6CO2(g) + 34H2O(l) + 30ATP4-

(aq)

(a) Calculate the ∆H° for the uncoupled oxidation of pyruvic acid toCO2 and H2O.

(b) Calculate the ∆H° for the coupled reaction and determine the

percentage of energy converted to ATP!.

Answer to 7.49:

(a) The separate equation for the uncoupled oxidation of pyruvic acid is:


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2C3H4O3(l) + 5O2(g) 6CO2(g) + 4H2O(l) ∆H°pyruv oxid "

!2##2 k$

∆H°pyruv oxid i% calculated from the heat% for formation a% %hown&

∆H°pyruv oxid " '(mol)!#*#.+ k$•mol!- )mol-)!2/+./ k$•mol!-0

! '2mol)!+/+./ k$•mol!-

" !2##2 k$

(b) 1ecall that ∆H°reaction" 2 k$•mol! for the formation of

one ATP4- from ADP3- and HOPO32-. Therefore, the energy required

to form 30 ATP4- will be 30 times as great.

30ADP3-(aq) + 30HOPO32-

(aq) + 30H3O+(aq) 60H2O(l) + 30ATP

4-(aq)

∆H°reaction " )#mol-)2 k$•mol!- " (# k$

The ∆H°coupled reaction i%& )!2##2 k$- (#k$ " !3#k$.

The 234 percent e5ciency i% calculated a% follow%&

Only 630kJ of the 2332kJ released in this oxidation are used effectively in the

production of ATP4-. The remaining 1703kJ is lost as heat.

Section 7.10. Pressure-Volume Work, Internal Energy, and Enthalpy

7.50. An inventor claims to have built a device that is able to do about 0.8 kJ of work

for every 1 kJ of thermal energy put into it. He is looking for investors to buy

shares in his company, so he can commercialize his invention. Would you invest

in his company? Explain why or why not.

Answer to 7.50:


34

4 e5ciency" (# $2##2 k$

6,"234


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You should, by all means, invest in this inventor (after you have seen proof that

his device can actually do what he says). This device is about 80% efficient in

converting random thermal motion to the directed motion of work. This is a good

deal better than the best engines now available. There is no violation of the first

law here, since he is not getting out more than he puts in. The “lost” energy

presumably remains in the device in some form or is expelled as thermal energy

to some other part of the surroundings. The inventors you have to look out for are

those who claim to get more work out of a thermal engine than the thermal energy

they put in; that is a violation of the first law.

7.51. The volume of a gas is decreased from 10. L to 1.0 L at a constant pressure of 5.0

atmospheres. Calculate the work associated with the process. Is work done on the

gas or by the gas? (1 L·atm = 101.3 J)

Answer to 7.51:

(a) P∆V = (5 atm) (10.0 L - 1.0 L) = 45 L·atm; 45 × 101.3 J

1 L ⋅ατµ= 4.6 × 103 J

(b) Work is being done on the gas, as the volume is decreasing from 10.0 L to 1.0

L. A positive sign is associated with work done on the system.

7.52. A hot air balloon is inflated by using a propane burner to heat the air in the

balloon. If during this process 1.5 × 108 J of heat energy cause the volume of the

balloon to change from 5.0 × 106 L to 5.5 × 106 L, what are the values of q, w, and

∆ E for this process? Assume that the balloon expands against a constant pressure

of 1.0 atmosphere. (1 L·atm = 101.3 J)

Answer to 7.52:

The value of q p is given, this is 1.5 × 108 J. This is the heat energy added

to the system at constant pressure.

The pressure-volume work done can be calculated in this manner.

w = P∆V

w = (1.0 atm) (5.5 × 106

L – 5.0 × 106

L)

w = (1.0 atm) (5.5 × 106

Λ 7 5.0 × 106 Λ) ×

101.3 ϑ

1 Λ8ατµ= 5× 10

7ϑ

Finally, ∆ E can be calculated from the values for q p and w; ∆ E = q p – P∆V

∆ E = (1.5 × 108 J) – (5 × 107 J) = 1.0 × 108 J


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7.53. Photographic flash bulbs, once more common than they are now, produced light

from the reaction of either zirconium or magnesium wire with oxygen in the bulb,

for example:

2Mg(s) + O2(g) → 2MgO(s) + energy (light and heat)

They became quite warm when used, but only rarely did they actually explode!

Discuss the changes in ∆ E , q, and w for the chemical reaction that takes place in a

flash bulb.

Answer to 7.53:

The chemical energy of reaction of the metal with oxygen is transformed to light

energy and heat energy in the course of the reaction. As long as the bulb does not

explode, this is a constant volume reaction. This means no pressure-volume work

is possible, so w = 0. This also means ∆ E = ∆ H .

7.54. A gaseous chemical reaction occurs in which the system loses heat and contracts

during the process. What are the signs, "+" or "-", for P∆V , ∆ E and ∆ H ? Explain

your reasoning.

Answer to 7.54:

Since ∆ H = ∆ E + P∆V , we know that ∆ H is negative since the system loses heat.

The system contracts so Vfinal < Vinitial. This means that P∆V will be negative since

P is always a positive number. ∆ E may be either positive or negative depending

on the magnitude of ∆ H and P∆V .

7.55. Consider two experiments in which a gas is compressed in a closed syringe by

pushing the plunger against the trapped gas. Both experiments use the same size

syringe.

Experiment A Experiment B


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(a) How does the pressure-volume work performed in Experiment A compare to

that performed in Experiment B? Assume the external pressure is the same in both

experiments. Explain your reasoning.

(b) Assume that a third experiment is performed using a syringe with a wider

diameter than the one used in the first two experiments. How would the pressure-

volume work compare with that in Experiments A if ∆ x is the same? Assume the

pressure is the same in both experiments.

Answer to 7.55:

(a) The pressure-volume work performed in Experiment A is less than that

performed in Experiment B. Given that the cross-sectional area is the same

for both syringes and the pressure is constant, the pressure-volume work is

directly proportional to the value of∆ x

, which is smaller for ExperimentA. The mathematical relationship is:

w = P × A × ∆ x

(b) Although ∆ x is the same for both experiments, ∆V is not the same because the

cross-sectional area of the syringe is greater in the third experiment. Because

pressure -volume work is a product of the pressure, which is constant, and the

change in volume, the work done in the third experiment is greater than that done

in Experiment A.

7.56. Consider the changes that occur in this reaction:

#H2(g) 92(g) → 29H#(g)

Is work done by the system (gaseous reactants and product) or on the system?

Explain your answer. Assume that this reaction occurs at constant temperature

and pressure.

Answer to 7.56:

We calculate ∆V by noting that 3 liters of H2(g) react with 1 liter of N2(g) to

produce 2 liters of NH3(g). In this case, ∆V = -2 liters and work is greater than

zero, indicating that work must enter the system.

7.57. Most of the nitrogen used to inflate an automobile airbag is produced by the

reaction of sodium azide:

2[Na+][N3–](s) —> 2Na(g) + 3N2(g)

A driver’s side airbag might typically contain about 95 g of sodium azide.


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(a) A mole of nitrogen gas occupies about 25 L at 25 °C. How large is the airbag

this nitrogen will inflate?

(b) How much work is done by the nitrogen as it inflates the airbag?

(c) If this exothermic reaction is investigated in a constant volume calorimeter,

will the measured thermal energy release be greater or less than the thermal

energy released at constant pressure? How large will the difference be? Clearly

explain your reasoning.

Answer to 7.57:

(a) The stoichiometric equation shows that 3 moles of nitrogen gas are

produced for every two moles of the azide that reacts. The azide has a

molar mass of 65 g. Therefore:

mol N2 formed =95 g azide

65 g ⋅µολ17

?

??

?

??

3 µολ Ν2

2 µολ αζιδε

?

?

?

?= 2.2 mol N2

The volume occupied by 2.2 mol N2 is 55 L (2.2 mol × 25 L·mol–1) so the

airbag is about 55 L in volume, which is approximately 2 cubic feet (or

you can imagine about 25 empty, two-liter soft drink bottles suddenly

appearing in front of you when the airbag deploys).

(b) The external pressure against which the airbag pushes to inflate is

atmospheric pressure, Patm, which is 1.0 × 105 Pa (N·m–2). The volume

change, ∆V , of the gas is 55 × 10–3 m3 (since we go from no gas to a final

volume of 55 L and a liter is 10–3 m3 = a cubic decimeter). Thus:

work done by the gas = Patm·∆V = (1.0 × 105 N·m–2)·(55 × 10–3 m3) = 5500 J

(c) We know that ∆ H = ∆ E +P·∆V and that the pressure-volume product is a

positive 5500 J. Therefore, the measured thermal energy release at constant

volume, qV = ∆ E , will be smaller by 5500 J than that measured at constant

pressure, qP = ∆ H .

7.58. For this reaction, as written, ∆ H ° = -484 kJ:

2H2(g) + O2(g) → 2H2O(g)When 0.5 mol of H2(g) was reacted with 0.3 mol of O2(g), at a constant pressure of

1.0 atm, the change in volume was –6.1 liters. Calculate how much work was

done and the value of ∆ E ° for this reaction. Is work done on the system or by the

system?

Answer to 7.58:


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Since ∆ E = ∆ H - P∆V , we will first need to calculate both ∆ H and P∆V to

obtain the value for ∆ E . Note that the initial amounts of H2(g) and O2(g)

are not in the 2:1 molar ratio demonstrated by the balanced equation. In

fact, H2(g) is the limiting reagent and is consumed completely while

0.05mol of O2(g) remains unreacted. ∆ H for the amounts given is:

:ork )!Pext ∆V - i% determined next&

9ote& :ork i% done on the %y%tem.

Therefore, ∆E " !2k$ .(2k$ " 2.k$

Section 7.11. What Enthalpy Doesn’t Tell Us

7.59. Indicate whether each of the following statements is true or false. If a statement is

false, write the correct statement.

(a) Consider a 10 mL sample of pure water at 25 °C and 1 atm pressure (state

A). The sample is cooled to 1 °C and then the pressure is reduced to 0.5 atm

(state B). It takes 5 hours to carry out the change from state A to state B. The

sample is then heated and the pressure is raised to 1 atm. In one minute the water

is back at 25 °C ( the sample is back at state A). The internal energy change in

going from state A to B is equal to, but opposite in sign to the internal energy

change going from state B to A.

(b) The work done in changing from state A to B and the work done in changing

from state B to A in part (a) are numerically the same but opposite in sign.

(c) The enthalpy change for a change of state of a system is independent of the

exact state of the reactants or products.

(d) At constant pressure the amount of heat absorbed or evolved by a system is

called the enthalpy change, ∆ H .


39

;H ".+mol H2)g- 6!/ $

2mol H2)g- "


40/53

(e) If volume does not change, the amount of heat released during a change of

state of a system is equal to the decrease in internal energy of that system.

(f) If a reaction is spontaneous, it is always exothermic.

(g) If the enthalpy change for this reaction, N2(g) + O2(g) → 2NO(g), is 180.5 kJ,

then the enthalpy change for this reaction, 1/2N2(g) + 1/2O2(g) → NO(g), is 90.2

kJ.

Answer to 7.59:

(a) True. Internal energy is a state function. The difference in the internal

energies of two states is independent of the path taken between the states.

(b) False. Work depends on the path, i.e., the method of going from one

state to another.

(c) False. The enthalpy of reaction is dependent on the exact state of the

reactants or products.

(d) True. The relationship between ∆ H and ∆ E is ∆ H = ∆ E +p∆V . At

constant volume the amount of heat absorbed or evolved by a system is

equal to eternal energy change.

(e) True. At constant volume ∆H = ∆E and qv= qp

(f) False. Endothermic reaction which are spontaneous include freezing

water at temperatures below 0oC and dissolving many salts (for example

NH4NO3).

(g) True.

7.60. When 1 mol of NH4Cl(s) is dissolved in water, forming NH4+(aq) and Cl-(aq), the

change in enthalpy is 14.8 kJ·mol-1. When 1 mol of CaCl2(s) is dissolved in water,

forming Ca2+(aq) and Cl-(aq), the change in enthalpy is –82.9 kJ·mol-1.

(a) :hich >eaker felt cool to touch after the %alt di%%olved in

water? :hich one felt warm?

(b) :hat would you have to do to maintain the temperature of

each >eaker at 2+°C?

(c) To determine ∆H° for di%%olution of %alt% in water, what

factor% mu%t you con%ider?

Answer to 7.60:

(a) The NH4Cl solution is cool to touch, while the CaCl2 solution is warm

to touch.


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(b) For the NH4Cl solution, 15.1 kJ of heat must be added. For the CaCl2

solution, 82.9 kJ must be removed.

(c) The concentration of the dissolved salt must be considered. The above values

for the enthalpy changes are obtained by measuring enthalpy changes and

extrapolating to the limit of infinite dilution. Also, we note that it is impossible to

assign the enthalpy changes associated with the individual ions of a salt without

giving an arbitrary enthalpy value to one of the ions.

7.61. Given that ∆ H °f for Cl-(aq) = –167.4 kJ·mol-1, use the enthalpy of reaction data

from Problem 7.60 to calculate ∆ H °f for NH4+(aq) and for Ca2+(aq).

Answer to 7.61:

For NH4+(aq):

∆H° form'9H)a@-0 ∆H° form'Cl!)a@-0 " ∆H° form'9HCl)%-0 +.

k$•mol!

∆H° form'9H)a@-0 )!(3. k$•mol!- " !#+. k$•mol! +.

k$•mol!

∆H° form'9H)a@-0 " !#2.* k$•mol!

or Ca

2

)a@-&

∆H° form'Ca2)a@-0 ∆H° form'2Cl

!)a@-0 " ∆H° form'CaCl2)%-0 ! /2.*

k$•mol!

∆H° form'Ca2)a@-0 ! ##./ k$•mol! " !3*+./ k$•mol! ! /2.*

k$•mol!

∆H° form'Ca2)a@-0 " !+#.* k$•mol!

7.62. A nineteenth-century chemist, Marcellin Berthelot, suggested that all chemical

processes that proceed spontaneously are exothermic. Is this correct? Give some

examples that justify your answer.

Answer to 7.62:


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Berthelot’s suggestion was incorrect. Endothermic processes that are spontaneous

include freezing water at temperatures below 0oC and dissolving many salts (for

example NH4NO3 ).

Section 7.13. EXTENSION — Ideal Gases and Thermodynamics

7.63. Consider two samples of gas in identical size containers. Which of these

statements is true? Explain your reasoning for each choice. Rewrite the false

statements to make them true.

(a) If the temperature of the two samples is the same, then the pressure of gas in

each container is the same.

(b) If the temperature and pressure of the two samples are the same, then the

number of moles of gas in each container is the same.

(c) If the gas in each container is the same and the temperature and number of

moles of gas in each container are the same, then the number of collisions with

the wall per unit time is the same in both containers.

Answer to 7.63:

(a) The statement, as given, is not true. Although the temperature and volume of

the gas samples are the same, we have no information about the quantity of gas,

that is, the number of moles of gas, in each container. The one with more moles

will have the higher pressure, since P = n( RT / V ) and all the quantities inside the

parentheses on the right are the same for both samples. We can amend the

statement in two ways: (1) If the temperature and number of moles of gas is thesame, then the pressure of gas in each container is the same or (2) If the

temperature of the two samples is the same, then the pressure will be higher in the

container with the greater number of moles of gas.

(b) This statement is true, since n = PV / RT and all the quantities on the right are

the same for both samples.

(c) This statement is true. To see why this is so, we note that the pressure of the

gas in each container is the same, since P = nRT / V and all the quantities on the

right are the same for both samples. Pressure is a result of collisions of the gas

molecules with the walls. Since the gas molecules an

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